Carcass wrote:
A student's score on the final test in a certain course was 60% greater than the average (arithmetic mean) score of the 2 other tests the student took in the course. The student's score on the final test was what percent of the student's average test score for the entire course?
A. \(33 \frac{1}{3} %\)
B. \(40%\)
C. \(75%\)
D. \(133 \frac{1}{3}%\)
E. \(160%\)
student's final test score: x1
avg of 2 other tests: (x2+x3)/2
The student's score on the final test was what percent of the student's average test score for the entire course?
x1/(avg of 3 tests)
.8(x2 + x3)
----------
.6(x2 + x3)
= 1.33 = 133%
x1 = 1.6 (x2+x3)/2
x1 = .8(x2+ x3)
the average of all three tests: (x1 + x2 + x3)/3
.8(x2 + x3) + x2 + x3
----------------------
3
1.8(x2 + x3)
-------------
3
= .6(x2 + x3)
The student's score on the final test was what percent of the student's average test score for the entire course?
Well you get that by dividing student's final score to the total average
.8(x2+x3)
----------
.6(x2 + x3)
= 130%